No, you're still missing the point :)

He's saying that SHA1'(m) = SHA1'(n) implies that SHA1"(m) = SHA1"(n).

Let's do this with a very simple, silly hash function.

Let h(x) = (x/2)%10 (in integer arithmetic, e.g, 5/2 = 2, so h(5) = (5/2)%10 = 2%10 = 2.) So then let h2(x) = h(h(x)) = (h(x)/2)%10, and h3(x) = h(h(h(x))) = (h2(x)/2) % 10.

Here's a table:

  x:  h  h2  h3
  0   0  0   0
  1   0  0   0
  2   1  0   0
  3   1  0   0
  4   2  1   0
  5   2  1   0
  6   3  1   0
  7   3  1   0
  8   4  2   1
  9   4  2   1
  10  5  2   1
  11  5  2   1
  12  6  3   1
  13  6  3   1
  14  7  3   1
  15  7  3   1
  16  8  4   2
  17  8  4   2
  18  9  4   2
  19  9  4   2
  20  0  0   0
  21  0  0   0
  ...

Hopefully this table shows that collisions are much more likely for h2 than for h, and more likely for h3 than for h2. The range of h3 is smaller than the range of h2 which is smaller than the range of h.